The correct curve between the magnetic induction $B$ along the axis of a long solenoid due to current flow $i$ in it and distance $x$ from one end is:

In a toroid,the magnetic field on the axis is to be calculated given the radius $r = 0.5 \, cm$,current $I = 1.5 \, A$,number of turns $N = 250$,and relative permeability $\mu_r = 700$. Find the magnetic field $B$ (in Tesla). (in $.5$)

$A$ toroidal solenoid with air core has an average radius $R$,number of turns $N$,and area of cross-section $A$. The self-inductance of the solenoid is (Neglect the field variation across the cross-section of the toroid).

$A$ long,straight wire of radius $a$ carries a current distributed uniformly over its cross-section. The ratio of the magnetic fields due to the wire at distance $\frac{a}{3}$ and $2a$ respectively from the axis of the wire is

$A$ solenoid has a core of a material with relative permeability $400$. The windings of the solenoid are insulated from the core and carry a current of $1 \text{ A}$. If the number of turns is $1000 \text{ per metre}$,the magnetic field $(B)$ is . . . . . . $\text{T}$. (Given: $\mu_0 = 4\pi \times 10^{-7} \text{ SI units}$)

The magnetic field intensity $(H)$ at the centre of a long solenoid having '$n$' turns per unit length and carrying a current $I$, when no material is kept in it is ($\mu_0$ = permeability of free space)